How do you solve Minimum Array End on LeetCode?

Minimum Array End (LeetCode #3133) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: The bitwise AND operation retains bits only where both operands have bits set.

What is the time complexity of Minimum Array End?

The optimal solution for Minimum Array End runs in O(1) time and uses O(1) space. This approach runs in constant time and space since it directly computes the last element without any loops.

What is the brute force solution for Minimum Array End?

The brute force approach for Minimum Array End is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible arrays of size n and checking if their bitwise AND equals x. This is straightforward but inefficient as it explores many combinations. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Minimum Array End?

Minimum Array End uses the following concepts: Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Array.

What are the interview tips for Minimum Array End?

Always consider edge cases, such as the smallest and largest values for n and x. Think about how bitwise operations work and how they can simplify your solution. Practice constructing arrays with specific properties to become comfortable with these types of problems.

What are common mistakes when solving Minimum Array End?

Failing to ensure that the array is strictly increasing. Overcomplicating the problem by trying to brute force instead of recognizing patterns.

#3133

Minimum Array End

Medium

Bit ManipulationBit ManipulationArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

The optimal approach constructs the array directly by leveraging the properties of bitwise operations. We can merge x with increasing integers to ensure the AND condition is satisfied while keeping the last element minimal.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array nums of size n.
2Step 2: Set nums[0] = x, then fill the rest of the array with x + i for i from 1 to n - 1.
3Step 3: Ensure that each element is greater than the previous one, which is guaranteed by the construction.

solution.py5 lines

1# Full working Python code
2
3def min_array_end_optimal(n, x):
4    return x + n - 1
5

ℹ

Complexity note: This approach runs in constant time and space since it directly computes the last element without any loops.

1The bitwise AND operation retains bits only where both operands have bits set.
2Constructing the array directly can save time by avoiding unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.